I posted a revision to my macro to reflect a new location of Column C for
you sequential numbers. Previously you said it was in column E. The
revision runs fine if the data is in C. The original runs fine if it is in
column E. Giving correct information improves your
chances of getting a working solution.

Please stay in the original thread so you don't waste people's time with
redundant solutions.

--
Regards,
Tom Ogilvy

"vernerv " wrote in message
...
Hey Bob,

Thanks for the code, but there's a problem.

The output looks like this:

166930.000000 5272038.000000 1 1.500000
165949.000000 5272484.000000 1 1.500000

165325.000000 5273019.000000 1 1.500000
164879.000000 5273733.000000 1 1.500000

164826.000000 5273945.000000 1 1.500000
163541.000000 5273019.000000 2 1.500000

163720.000000 5274804.000000 2 1.500000
163987.000000 5276409.000000 2 1.500000

But I would like it like this:

166930.000000 5272038.000000 1 1.500000
165949.000000 5272484.000000 1 1.500000
165325.000000 5273019.000000 1 1.500000
164879.000000 5273733.000000 1 1.500000
164826.000000 5273945.000000 1 1.500000

163541.000000 5273019.000000 2 1.500000
163720.000000 5274804.000000 2 1.500000
163987.000000 5276409.000000 2 1.500000


The number after 1,2.....till 379. I would need to have a line breaks
after all the 1's, the 2's and like that till the last one, ie. 379

Any suggestion?

Cheers
Elvis


---
Message posted from http://www.ExcelForum.com/